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3 years ago

Why does stress build up at transform boundaries?

1 answer:

3 0

Tectonic plate edges are not smooth. At transform boundaries, plates move against each other, but since the plates aren't smooth, lockups form. Stress is built up as both sides continue to steadily push against each other until there is sudden movement, where both plates quickly slide until they lock up again. This is what causes earthquakes.

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Find f0, the magnitude of the upward force on the table due to the leg at (0, 0.

Katena32 [7]

Assuming that the table has 4 legs. the upward force is equal to the downwardd force. the downward force of the table is the weight of the table. so the upward force is also the weight of the table but in opposite direction, so the upward force in on one leg is 1/4 of the weight of the table

6 0

4 years ago

A single-turn circular loop of radius 14 cm is to produce a field at its center that will just cancel the earth's magnetic field

Olin [163]

Answer:

49 A

Explanation:

We are given that

Number of turns,N=1

Radius,r=14 cm= $\frac{14}{100}=0.14 m$

1 m=100 cm

Magnetic field,B=0.7 G= $0.7\times 10^{-4} T$

$1G=10^{-4} T$

$\mu_0=4\pi\times 10^{-7} Tm/A$

We know that magnetic field

$B=\frac{\mu_0IN}{2\pi r}$

Substitute the values

$0.7\times 10^{-4}=\frac{4\pi\times 10^{-7}\times 1\times I}{2\pi\times 0.14}$

$I=\frac{0.7\times 10^{-4}\times 2\pi\times 0.14}{4\pi\times 10^{-7}\times 1}$

$I=49A$

6 0

3 years ago

A block of mass 10 kg moves from position A to position B shown in the figure above. The speed of the block is 10 m/s at A and 4

Otrada [13]

We have that the block is moving horizontally. Hence, its potential energy due to gravity stays the same. The only change in its mechanical energy is the one due to the change of speed. This reduction of its kinetic energy, due to the conservation of energy, is equal to the work that friction does. We have that at A the kinetic energy is : K=1/2*m*u^2=10*10*10/2=500J. At B, we have that K=1/2*10*16=80J. Sine we have that the initial value is 500, the work from the friction force (opposite to the movement of the object) is 80-500=420J.

7 0

4 years ago

Consider a spherical Gaussian surface of radius R centered at the origin. A charge Q is placed. inside the sphere. Where should

Elanso [62]

I think you forgot to give the options along with the question. I am answering the question based on my knowledge and research. "<span>The charge can be located anywhere, since flux does not depend on the position of the charge as long as it is inside the sphere" gives the answer to the question. I hope that this answer has come to your help.</span>

5 0

3 years ago

Consider a uniformly wound solenoid having N=210 turns, length l=0.18 m, and cross-sectional area A = 4.00 cm2. Assume l is much

strojnjashka [21]

Answer:

Explanation:

Number of turns

N = 210turns

Length of solenoid

l = 0.18m

Cross sectional area

A = 4cm² = 4 × 10^-4m²

A. Inductance L?

Inductance can be determined using

L = N²μA/l

Where

μ is a constant of permeability of the core

μ = 4π × 10^-7 Tm/A

A is cross sectional area

l is length of coil

L is inductance

Therefore

L = N²μA / l

L=210² × 4π × 10^-7 × 4 × 10^-4 / 0.18

L = 1.23 × 10^-4 H

L = 0.123 mH

B. Self induce EMF ε?

EMF is given as

ε = -Ldi/dt

Since rate of decrease of current is 120 A/s

Then, di/dt = —120A/s, since the current is decreasing

Then,

ε = -Ldi/dt

ε = - 1.23 × 10^-4 × -120

ε = 0.01478 V

ε ≈ 0.015 V

4 0

3 years ago